How do you solve #log_2(x+3)+log_2(x-3) = 4#?

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Answer 1 · 2018-04-05T22:29:46

Let's use one of the log laws, which says that #log_b(a) + log_b(c)# can be rewritten as #log_b(a*c)#

#log_2(x+3)+log)2(x-3) = 4#

#log_2((x+3)(x-3)) = 4#

#log_2(x^2-9) = 4#

We can write a log , such as #log_b(a) = c# to be #b^c = a# (note, the base remains the same)

#2^4 = x^2 - 9#

#16 = x^2 - 9#

#25 = x^2#

#x = 5#